Examples to be solved by Elimination by Comparison.

EXAMPLES.

1. To solve any two equations of the first degree with two unknown quantities.

Solution. These equations may, as in art. 110, be reduced to the forms

The values of x, obtained from these equations, are

which, being placed equal to each other, give

being the same values as those obtained in art. 110.

are

Examples to be solved by Elimination by Comparison.

Solution. The values of ́x, obtained from these equations,

the first of which being placed equal to each of the others gives, by reduction,

z = 4, y = 3;

whence we get, from either value of x, by substitution,

7. A person has two horses, and two saddles one of which cost $50, the other $2. If he places the best saddle upon the first horse, and the worst upon the second, then the latter is worth $8 less than the other; but if he puts the worst saddle upon the first horse, and the best upon the other, then the latter is worth 32 times as much as the first. What is the value of each horse?

Ans. The first $30, the second $70.

8. What fraction is that, whose numerator being doubled, and denominator increased by 7, the value becomes; but the denominator being doubled, and the numerator increased by 2, the value becomes &?

Ans.

9. A wine merchant has two kinds of wine. If he mix 3 gallons of the worst with 5 of the best, the mixture is worth $1 per gallon; but, if he mix 34 gallons of the worst with 83 gallons of the best, the mixture is worth $1,033 per gallon. What does each wine cost per gallon?

Ans. The best $1,12, the worst $0,80.

10. A wine merchant has two kinds of wine. If he mix a gallons of the first with b gallons of the second, the mixture is worth c dollars per gallon; but, if he mix a' gallons of the first with b' gallons of the second, the mixture is worth c' dollars per gallon. What does each wine cost per

gallon?

11. Three masons, A, B, C, are to build a wall. A and B, jointly, could build this wall in 12 days; B and C could accomplish it in 20 days; A and C would do it in 15 days. What time would each take to do it alone?

Ans. A requires 20, B 30, and C 60 days.

12. Three laborers are employed in a certain work. A and B would, together, complete it in a days; A and C require b days; B and C require c days. In what time would each accomplish it singly?

13. A cistern may be filled by three pipes, A, B, C. By the pipes A and B, it could be filled in 70 minutes; by the pipes A and C, in 84 minutes; and by the pipes B and C, in 140 minutes. In what time would each pipe fill it?

Ans. A in 105, B in 210, and C in 420 minutes.

14. A, B, C play faro. In the first game A has the bank, B and C stake the third part of their money, and win. In the second game B has the bank, A and C stake the

Then C takes the

third part of their money and also win. bank, A and B stake the third part of their money and also win. After this third game they count their money, and find that they have all the same sum of 64 ducats. much had each when they began to play?

How

Ans. A had 75, B 63, and C 54 ducats.

Elimination by Addition and Subtraction.

15. Five friends, A, B, C, D, E, jointly spend $879 at an inn. This sum is to be paid by one of them; but on counting the dollars they had in their pockets, they find that none of them had, alone, enough for this purpose. If, then, one of them is to pay it, the others must give him a part of their money. A can pay, if he receives one fourth; B, if he receives one fifth; C, if he receives one sixth; D, if he receives one seventh; and E, if he receives one eighth of the others' money. How much has each?

Ans. A $319, B $459, C 543, D $599, E $639.

116. Third Method of solving the Problem of art. 106, called that of Elimination by Addition and Sub

traction.

Solution. This method is generally inapplicable to trancendental equations, but otherwise, to equations of any degree whatever; that is, it is a method which can be successfully applied in all such cases to eliminate one unknown quantity after another, until the given equations are reduced

to one.

In order to eliminate an unknown quantity from two equations which contain it, reduce them as in arts. 79 and 88, and arrange their terms according to the powers of the quantity to be eliminated, taking out each power as a factor from the terms which contain it.

It being now recollected that the second member of each of these equations is zero, it will appear evident that, if the first members are divided one by the other, the remainder arising from this division must likewise be equal to zero; for this remainder is the difference between the dividend and a certain multiple of the divisor, that is, between zero and a certain multiple of zero.